Why is it not possible to obtain F2 by electrolysis of aqueous NaF, aqueous HF or anhydrous.
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For aqueous solutions of fluorine compounds, you will always end up electrolyzing the water instead. You can see this by comparing reduction potentials:
O2 + 4H+ + 4e- -----> 2H2O (+1.23 V)
F2 + 2e- -----> 2F- (+2.87 V)
These are written in the reduction direction; as oxidation reactions the voltages are flipped, and it's much easier to oxidize water than to oxidize fluoride ion.
You can also notice that even if electrolysis could produce fluorine gas, it would react with water; the redox reaction obtained by combining the above two half-reactions is highly favourable:
2F2 + 2H2O -----> 4HF + O2
Anhydrous HF does not work because it is a poor conductor of electricity (much like distilled water). When Henri Moissan first isolated fluorine, he used a mixture of potassium bifluoride and hydrogen fluoride. The former is an ionic compound so it causes the solution to become a good conductor.
O2 + 4H+ + 4e- -----> 2H2O (+1.23 V)
F2 + 2e- -----> 2F- (+2.87 V)
These are written in the reduction direction; as oxidation reactions the voltages are flipped, and it's much easier to oxidize water than to oxidize fluoride ion.
You can also notice that even if electrolysis could produce fluorine gas, it would react with water; the redox reaction obtained by combining the above two half-reactions is highly favourable:
2F2 + 2H2O -----> 4HF + O2
Anhydrous HF does not work because it is a poor conductor of electricity (much like distilled water). When Henri Moissan first isolated fluorine, he used a mixture of potassium bifluoride and hydrogen fluoride. The former is an ionic compound so it causes the solution to become a good conductor.
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